Solve Equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ in Real Numbers

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Numbers Real numbers
Click For Summary
SUMMARY

The equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$ has been conclusively solved, with $a=0$ identified as the sole real solution. The discussion highlighted the derivation from the equation $3a = 3\sqrt[3]{a(a^2-1)}$, leading to the simplification $a^3 = a(a^2-1)$. Participants confirmed that $a=0$ is indeed the only valid solution after addressing calculation errors in alternative methods.

PREREQUISITES
  • Understanding of cube roots and their properties
  • Familiarity with algebraic manipulation of equations
  • Knowledge of real number solutions in polynomial equations
  • Basic skills in solving equations involving radicals
NEXT STEPS
  • Study the properties of cube roots in depth
  • Explore polynomial equations and their solutions
  • Learn about methods for verifying solutions in algebra
  • Investigate alternative approaches to solving radical equations
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in solving polynomial equations involving radicals.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$
 
Last edited:
Mathematics news on Phys.org
anemone said:
Solve in real numbers the equation $\sqrt[3]{a-1}+\sqrt[3]{a}+\sqrt[3]{a+1}=0$

using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
 
kaliprasad said:
using

x+y+z=0=>$x^3+y^3+x^3 = 3xyz$

we get

$(3a)^3 = 3a(a^2-1)$
a= 0 or +/-$\sqrt(1/8)$
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]
 
Opalg said:
Neat method, but I get a different answer.

[sp]From $3a = 3\sqrt[3]{a(a^2-1)}$, I get $a^3 = a(a^2-1)$, with $a=0$ the only solution.[/sp]

There was a calculation mistake in my method
 
Thanks to both of you for participating and yes, $a=0$ is the only answer to the problem.:)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K